Subtopic: Preservation of food by cooling and freezing
Question Statement:
In a cold storage, 10 metric ton of potato is to be brought down from 30 to 8 °C storage temperature in 6 hours of air blast at the evaporator temperature of –10 °C. Specific heat capacity of potato is 3.2 kJ kg⁻¹ K⁻¹. COP of the refrigeration cycle deployed is 4.2 with evaporator load extraction capacity of 210 kJ kg⁻¹. Neglecting the respiration load of potato, the refrigerant flow rate and compressor power requirement will be
Options:
(A) 3.9 kg min⁻¹; 7.76 kW
(B) 9.3 kg min⁻¹; 8.50 kW
(C) 9.3 kg min⁻¹; 7.76 kW
(D) 3.9 kg min⁻¹; 8.50 kW
Official Answer Key:
C
Step-by-Step Solution & Explanation
• Total heat to be removed = m Cp ΔT = 10,000 kg × 3.2 kJ/kg·K × (30–8)°C = 10,000×3.2×22 = 704,000 kJ.
• Cooling load rate = 704,000 kJ / 6 h = 117,333.3 kJ/h = 32.59 kJ/s = 32.59 kW.
• Refrigerant flow rate = cooling load / evaporator load extraction capacity = 32.59 kW ÷ 210 kJ/kg = 0.1552 kg/s = 0.1552×60 ≈ 9.31 kg/min.
• Compressor power = cooling load / COP = 32.59/4.2 ≈ 7.76 kW.
• Final Answer: (C) 9.3 kg min⁻¹; 7.76 kW
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